Finding velocity as a function of position, constant acceleration

[crowned]

This isn't actually a homework question, it's for a vehicle safety project at work, but I think it fits in this forum.

Homework Statement

A truck is traveling down the road at 60 mph, and begins to decelerate at a rate of 0.8 g.

I'd like to find the total distance it takes the truck to stop, and come up with a formula for the velocity as a function of displacement from the original braking point.

Homework Equations

I think that the relevant equation is v² = 2a(s₂-s₁)+v₁²

The Attempt at a Solution

Well, I know the truck is traveling at 88 fps, and slowing at a rate of 0.8 * 32fps², so it will take

88 fps / (32.2 fps² * 0.8 g) * 44 fps = 150 feet to stop

But when I plug those numbers into the above equation, I get

v₂ = (2 * (0.8 * 32.2 * 60 / 88) * (-x) + (60)²)^0.5

That has the right y-intercept (65 mph), but the wrong x-intercept, around 104'

If I change the "2" on the right side of the equation to 1.38, it looks right, but I can't work out why it would be 1.38 instead of two.

Argh! Any help much appreciated!

 

[PeroK]

The time it takes a truck to stop is v/a. The average speed during this time is v/2. So the distance traveled when the truck stops is [tex]s = v^2/2a[/tex]
And, if the truck is still traveling at speed v1, then we have:
[tex]s_1 = (v^2-v_1^2)/2a[/tex]
Where a is the magnitude of the deceleration and v is the initial speed.

 

[crowned]

The time it takes a truck to stop is v/a. The average speed during this time is v/2. So the distance traveled when the truck stops is [tex]s = v^2/2a[/tex]
And, if the truck is still traveling at speed v1, then we have:
[tex]s_1 = (v^2-v_1^2)/2a[/tex]
Where a is the magnitude of the deceleration and v is the initial speed.

Thanks for your response. However, I understand that part already--what I'm trying to work out is the equation for velocity as a function of position.

 

[PeroK]

You mean you understood all that but yet you couldn't rearrange the equation for yourself:

[tex]s_1 = (v^2-v_1^2)/2a[/tex]
[tex]v_1^2 = v^2 - 2as_1[/tex]
[tex]v_1 = \sqrt{v^2 - 2as_1}[/tex]

 

[crowned]

Hi--thanks for the response. Well, I worked my way to the final equation you posted (it matches up to the final equation in my original post--may not have been clear, I didn't realize I could write in LaTeX). The problem is that I calculate a stopping distance of 150' by using average speed * time, but a "displacement when velocity equals zero" of 120' when I use the last of the equations you just posted.

Maybe I'm doing something else wrong? Here's the equation I'm using, with variables.

[tex]v_1 = \sqrt{(65mph)^2 - 2(0.8 g \cdot 32.2fps^2 \cdot 60mph/80fps) \cdot x}[/tex]

 

[PeroK]

Are you trying to find speed as a function of position, or initial speed as a function of total stopping distance?

 

[crowned]

Are you trying to find speed as a function of position, or initial speed as a function of total stopping distance?

Speed as a function of position. I'd like to know how fast the truck is going at a given displacement.

 

[PeroK]

You need all your quantities expressed in the same units: fps. Do that before you plug anything into your equation.

 

[crowned]

You need all your quantities expressed in the same units: fps. Do that before you plug anything into your equation.

Well, everything is expressed in mph.

 

[crowned]

Oh my gosh, how did that work!

Expressed in fps, it works fine.

y-intercept 88 fps, x intercept 150'

So to change the y-axis to mph, I just multiply the right side of the equation by 60mph/88fps, right?

Sheesh, didn't see that one coming.

 

[PeroK]

1 mph = 5280/3600 fps = 88/60 fps, so yes.

 

[crowned]

Thanks so much for your help on this. Graphs are coming out perfectly.